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Marysya12 [62]
4 years ago
5

Suppose a homeless shelter provides meals and sleeping cots to those in need. A rectangular cot measures 6 feet long by 3 ½ feet

wide. Find the cot's diagonal distance from corner to corner. Round your answer to the nearest hundredth foot. 6.95 feet 9.64 feet 9.65 feet 6.94 feet
Mathematics
1 answer:
hjlf4 years ago
6 0

Answer:

6.95 feet

Step-by-step explanation:

The shape of the cot is rectangular. A diagonal of the rectangle divides the rectangle into two Congruent Right Angled triangles. The length and width of the rectangle become the legs of the right triangle and the diagonal is the hypotenuse of the right triangle.

In order to find the length of the hypotenuse which is the diagonal in this case we can use the Pythagoras Theorem. According to the theorem, square of hypotenuse is equal to the sum of square of its legs. So for the given case, the formula will be:

\textrm{(Diagonal)}^{2}=\textrm{(Length)}^{2}+\textrm{(Width)}^{2}\\\\ \textrm{(Diagonal)}^{2}=6^{2}+3.5^{2}\\\\ \textrm{(Diagonal)}^{2}=48.25\\\\ \textrm{(Diagonal)}=\sqrt{48.25}=6.95

Thus, rounded of to nearest hundredth foot, the diagonal distance from corner to corner is 6.95 feet

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The answers to this question
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For questions, 1-2, determine whether experiment is a binomial experiment. If it is, identify a success, specify the values of n
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Answer:

Part 1

X="number of U.S. households that own a dedicated game console"

Is a binomial experiment we an event defined with the associated probability and we have specific trials.

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=8, p=0.49)

Y= "number of cards that are hearts."

Thats not a binomial experiment since the probability for each trial changes since we are doing the experiment without replacment

Part 2

a) P(X=2)=(6C2)(0.39)^2 (1-0.39)^{6-2}=0.316

b) P(X \geq 5) = P(X=5)+P(X=6)

P(X=5)=(6C5)(0.39)^5 (1-0.39)^{6-5}=0.033

P(X=6)=(6C6)(0.39)^6 (1-0.39)^{6-6}=0.00352

P(X \geq 5) = P(X=5)+P(X=6)=0.033+0.00352=0.0365

Part 3

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=6, p=0.34)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Part 4

E(X)=np=4*0.69=2.76

The variance \sigma^2 = np(1-p) = 4*0.69*(1-0.69) =0.8556

\sigma=\sqrt{np(1-p)}=\sqrt{4*0.69(1-0.69)}=0.925

Unusual outcomes would be considered values above or below 2 deviations from the mean for example 2.76-(2*0.925) =0.91 or 2.76+2(0.925)=4.61[/tex]. X=0 and X=4 would be considered as unusual values.

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Part 1

X="number of U.S. households that own a dedicated game console"

Is a binomial experiment we an event defined with the associated probability and we have specific trials.

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=8, p=0.49)

Y= "number of cards that are hearts."

Thats not a binomial experiment since the probability for each trial changes since we are doing the experiment without replacment

Part 2

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=6, p=0.39)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

And we want to find this probability:

a) P(X=2)=(6C2)(0.39)^2 (1-0.39)^{6-2}=0.316

b) P(X \geq 5) = P(X=5)+P(X=6)

P(X=5)=(6C5)(0.39)^5 (1-0.39)^{6-5}=0.033

P(X=6)=(6C6)(0.39)^6 (1-0.39)^{6-6}=0.00352

P(X \geq 5) = P(X=5)+P(X=6)=0.033+0.00352=0.0365

Part 3

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=6, p=0.34)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Part 4

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=4, p=0.69)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

The expected value is given by:

E(X)=np=4*0.69=2.76

The variance \sigma^2 = np(1-p) = 4*0.69*(1-0.69) =0.8556

\sigma=\sqrt{np(1-p)}=\sqrt{4*0.69(1-0.69)}=0.925

Unusual outcomes would be considered values above or below 2 deviations from the mean for example 2.76-(2*0.925) =0.91 or 2.76+2(0.925)=4.61[/tex]. X=0 and X=4 would be considered as unusual values.

6 0
4 years ago
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